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Question Number 15955 by Tinkutara last updated on 16/Jun/17

The motion of a particle moving along x-axis is represented by the equation (dv/dt) = 6 − 3v, where v is in m/s and t is in second. If the particle is at rest at t = 0, then (1) The speed of the particle is 2 m/s when the acceleration of particle is zero (2) After a long time the particle moves with a constant velocity of 2 m/s (3) The speed is 0.1 m/s, when the acceleration is half of its initial value (4) The magnitude of final acceleration is 6 m/s^2

$$\mathrm{The}\:\mathrm{motion}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{moving}\:\mathrm{along} \\ $$$${x}-\mathrm{axis}\:\mathrm{is}\:\mathrm{represented}\:\mathrm{by}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\frac{{dv}}{{dt}}\:=\:\mathrm{6}\:−\:\mathrm{3}{v},\:\mathrm{where}\:{v}\:\mathrm{is}\:\mathrm{in}\:\mathrm{m}/\mathrm{s}\:\mathrm{and}\:{t}\:\mathrm{is} \\ $$$$\mathrm{in}\:\mathrm{second}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{at}\:\mathrm{rest}\:\mathrm{at}\:{t}\:= \\ $$$$\mathrm{0},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{The}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{2}\:\mathrm{m}/\mathrm{s} \\ $$$$\mathrm{when}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{particle}\:\mathrm{is} \\ $$$$\mathrm{zero} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{After}\:\mathrm{a}\:\mathrm{long}\:\mathrm{time}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{moves} \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{2}\:\mathrm{m}/\mathrm{s} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{The}\:\mathrm{speed}\:\mathrm{is}\:\mathrm{0}.\mathrm{1}\:\mathrm{m}/\mathrm{s},\:\mathrm{when}\:\mathrm{the} \\ $$$$\mathrm{acceleration}\:\mathrm{is}\:\mathrm{half}\:\mathrm{of}\:\mathrm{its}\:\mathrm{initial}\:\mathrm{value} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{The}\:\mathrm{magnitude}\:\mathrm{of}\:\mathrm{final}\:\mathrm{acceleration} \\ $$$$\mathrm{is}\:\mathrm{6}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \\ $$

Commented by prakash jain last updated on 16/Jun/17

∫(dv/(6−3v))=∫dt −(1/3)ln (6−3v)=t+C t=0,v=0 −(1/3)ln 6=C (1/3)ln ((6−3v)/6)=−t ((6−3v)/6)=e^(−3t) ⇒v=2(1−3e^(−t) ) ...(A) a=0⇒(dv/dt)=0⇒6−3v=0⇒v=2 t→∞,e^(−3t) →0⇒v=2 (from A) t→∞,v→z2,(dv/dt)→0 t=0,v=0⇒a_0 =6−3v=6 a_t =3⇒v=1

$$\int\frac{{dv}}{\mathrm{6}−\mathrm{3}{v}}=\int{dt} \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left(\mathrm{6}−\mathrm{3}{v}\right)={t}+{C} \\ $$$${t}=\mathrm{0},{v}=\mathrm{0} \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\mathrm{6}={C} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\frac{\mathrm{6}−\mathrm{3}{v}}{\mathrm{6}}=−{t} \\ $$$$\frac{\mathrm{6}−\mathrm{3}{v}}{\mathrm{6}}={e}^{−\mathrm{3}{t}} \Rightarrow{v}=\mathrm{2}\left(\mathrm{1}−\mathrm{3}{e}^{−{t}} \right)\:\:\:...\left({A}\right) \\ $$$${a}=\mathrm{0}\Rightarrow\frac{{dv}}{{dt}}=\mathrm{0}\Rightarrow\mathrm{6}−\mathrm{3}{v}=\mathrm{0}\Rightarrow{v}=\mathrm{2} \\ $$$${t}\rightarrow\infty,{e}^{−\mathrm{3}{t}} \rightarrow\mathrm{0}\Rightarrow{v}=\mathrm{2}\:\left({from}\:{A}\right) \\ $$$${t}\rightarrow\infty,{v}\rightarrow{z}\mathrm{2},\frac{{dv}}{{dt}}\rightarrow\mathrm{0} \\ $$$${t}=\mathrm{0},{v}=\mathrm{0}\Rightarrow{a}_{\mathrm{0}} =\mathrm{6}−\mathrm{3}{v}=\mathrm{6} \\ $$$${a}_{{t}} =\mathrm{3}\Rightarrow{v}=\mathrm{1} \\ $$

Commented by Tinkutara last updated on 16/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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